The code you needs depends on what you mean by "an empty space".

• If you mean the ASCII / Latin-1 / Unicode space character (0x20) aka SP, then:

  if (ch == ' ') {
      // ...
  }
  

• If you mean any of the traditional ASCII whitespace characters (SP, HT, VT, CR, NL), then:

  if (ch == ' ' || ch == '\t' || ch == '\r' || ch == '\n' || ch == '\x0b') {
      // ...
  }
  

• If you mean any Unicode whitespace character, then:

  if (Character.isWhitespace(ch)) {
      // ...
  }
  

Note that there are Unicode whitespace includes additional ASCII control codes, and some other Unicode characters in higher code planes; see the javadoc for Character.isWhitespace(char).

What you wrote was this:

    if (Equals(ch, " ")) {
        // ...
    }

This is wrong on a number of levels. Firstly, the way that the Java compiler tries to interpret that is as a call to a method with a signature of boolean Equals(char, String).

• This is wrong because no method exists, as the compiler reported in the error message.
• Equals wouldn't normally be the name of a method anyway. The Java convention is that method names start with a lower case letter.
• Your code (as written) was trying to compare a character and a String, but char and String are not comparable and cannot be cast to a common base type.

There is such a thing as a Comparator in Java, but it is an interface not a method, and it is declared like this:

    public interface Comparator<T> {
        public int compare(T v1, T v2);
    }

In other words, the method name is compare (not Equals), it returns an integer (not a boolean), and it compares two values that can be promoted to the type given by the type parameter.

Someone (in a deleted Answer!) said they tried this:

    if (c == " ")

That fails for two reasons:

• " " is a String literal and not a character literal, and Java does not allow direct comparison of String and char values.

• You should NEVER compare Strings or String literals using ==. The == operator on a reference type compares object identity, not object value. In the case of String it is common to have different objects with different identity and the same value. An == test will often give the wrong answer ... from the perspective of what you are trying to do here.

boolean containsWhitespace = false;
for (int i = 0; i < text.length() && !containsWhitespace; i++) {
    if (Character.isWhitespace(text.charAt(i)) {
        containsWhitespace = true;
    }
}
return containsWhitespace;

or, using Guava,

boolean containsWhitespace = CharMatcher.WHITESPACE.matchesAnyOf(text);

A space is a character similar to any other letter:

char space[] = {' '};

or

myCharArray[i] = ' ';